Activities of Enzymes

William Tivol tivol at news.wadsworth.org
Tue Aug 19 15:14:18 EST 1997


Tham Seng Choe (sengchoe at pobox.org.sg) wrote:

Dear Tham,

: I have some queries regarding the workings of an enzyme.

: We understand from most books that the enzyme speeds up a reaction BUT does
: not change the equilibrium of the reaction. Assuming a general equation ...

: S + Enzyme(E) ---> ES-complex ---> E + P

: An enzyme speeds up the formation of a product by providing an alternative
: pathway with lower activation energy so that the forward reaction proceeds
: with a faster rate. However, this does not show how the corressponding
: backward reaction rate is adjusted (increased in this case) to keep the
: equilibrium of the entire reaction unchanged.

	If you plot the free energy of the system along the reaction 
coordinate--substrate(s) to the left, intermediate(s) in the middle & 
product(s) on the right, you will get something like
                                 ________
                                /        \
                               /          \
                              /            \___________
                    _________/              

for a (endothermic) reaction in the absence of an enzyme, and
                               __________
                              /          \__________
                    _________/      

in the presence of the enzyme.  The equilibrium constant is determined by
the difference in the free energies, and is, as you say, unchanged.  The
lower free energy of the intermediate state(s) has the same effect on both
the forward and backward rates, since delta-delta-G (the change in free en-
ergy going from substrate to intermediate in the absence of enzyme minus that
in the presence of enzyme) is the same in each direction.

: Does the enzyme catalyse the backward reaction as well?

	Yes.  This has been shown experimentally.  At all concentrations
of reactants and products both the forward and backward reactions occur.
Usually, there is an excess of substrate, and the product is removed by
some subsequent step (reaction, transport, etc.), which causes the number
of backward reactions (=rate*[product(s)]) to be negligable, but if there
is little or no substrate and an excess of product, then the number of
backward reactions will be >> that of forward reactions.

: If so, would this
: not compromise on the "specificity" of the enzyme?

	No.  The enzyme would still form specific "product(s)"--the initial
substrate(s)--from "substrate(s)"--the initial products.

: And would this not
: provide a form a competitive inhibition which would decrease the efficiency
: of the enzyme?

	No.  However, some enzymes are regulated by the presence of product;
this can be allosteric inhibition.

: So, how is the increase in the rate of the forward reaction balanced to
: maintain the equilibrium?

	The forward rate is determined by the free energy difference between
the initial state and the highest intermediate state, that of the backward
reaction is likewise determined by delta-G between the final state and the
highest intermediate state.  These rates have the same ratio--the equilib-
rium constant--regardless of the energy of the highest intermediate.
				Yours,
				Bill Tivol



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